Let point $O$ be the origin of a three-dimensional coordinate system, and let points $A,$ $B,$ and $C$ be located on the positive $x,$ $y,$ and $z$ axes, respectively.  If $OA = \sqrt[4]{75}$ and $\angle BAC = 30^\circ,$ then compute the area of triangle $ABC.$
Answer: Let $b = OB$ and $c = OC.$

[asy]
import three;

size(250);
currentprojection = perspective(6,3,2);

triple A, B, C, O;

A = (3,0,0);
B = (0,4,0);
C = (0,0,2);
O = (0,0,0);

draw(O--(5,0,0));
draw(O--(0,5,0));
draw(O--(0,0,3));
draw(A--B--C--cycle);

label("$A$", A, S);
label("$B$", B, S);
label("$C$", C, NW);
label("$O$", O, S);
label("$b$", (O + B)/2, N);
label("$c$", (O + C)/2, E);
[/asy]

By the Law of Cosines on triangle $ABC,$
\begin{align*}
BC^2 &= AB^2 + AC^2 - 2 \cdot AC \cdot AB \cos \angle BAC \\
&= AC^2 + AB^2 - AB \cdot AC \sqrt{3}.
\end{align*}From Pythagoras,
\[b^2 + c^2 = c^2 + \sqrt{75} + b^2 + \sqrt{75} - AB \cdot AC \sqrt{3},\]which gives us $AB \cdot AC = 10.$

Then the area of triangle $ABC$ is
\[\frac{1}{2} \cdot AB \cdot AC \sin \angle BAC = \frac{1}{2} \cdot 10 \cdot \frac{1}{2} = \boxed{\frac{5}{2}}.\]